Light OJ 1214 Large Division （大数取模）
题目：
Given two integers, a and b, you should check whether a is divisible by b&nbs***ot. We know that an integer a is divisible by an integer b if and only if there exists an integer c such that a = b * c.
Input
Input starts with an integer T (≤ 525), denoting the number of test cases.
Each case starts with a line containing two integers a (-10^200 ≤ a ≤ 10^200) and b (|b| > 0, b fits into a 32 bit signed integer). Numbers will not contain leading zeroes.
Output
For each case, print the case number first. Then print 'divisible' if a is divisible by b. Otherwise print 'not divisible'.
Sample Input
6
101 101
0 67
-101 101
7678123668327637674887634 101
11010000000000000000 256
-202202202202000202202202 -101
Sample Output
Case 1: divisible
Case 2: divisible
Case 3: divisible
Case 4: not divisible
Case 5: divisible
Case 6: divisible
注：需要判断输入数字字符串a的首位为‘-’的情况和b为负数的情况。
代码：
#include <iostream>
#include <cstring>
using namespace std;
char a[100000];
int main()
{
long long t,i,k,b,n,h;
cin>>t;
h=1;
while(t--)
{
cin>>a;
cin>>b;
if(b<0) b=-b;
n=strlen(a);
if(a[0]=='-')
{
for(i=0;i<n-1;i++)
{
a[i]=a[i+1];
}
n--;
}
k=0;
for(i=0;i<n;i++)
{
k=(k*10+a[i]-'0')%b;
}
cout<<"Case "<<h++<<": ";
if(!k) cout<<"divisible"<<endl;
else cout<<"not divisible"<<endl;
}
return 0;
}
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